%%
%% lecture16.tex
%% 
%% Made by alex
%% Login   <alex@tomato>
%% 
%% Started on  Sat Dec 17 12:22:56 2011 alex
%% Last update Sat Dec 17 12:22:56 2011 alex
%%

We will go to the $K$-theory; before that, we would like to prove
a simple lemma.

\begin{ddanger}
Take a complex vector bundle $(E,B,F,p)$ considered as a $\U{n}$
bundle (since $\GL{n,\CC}\homotopic\U{n}$ homotopic, so we
consider homotopy classifications). Introduce the Hermitian
product on every fibre, pick a bases $e_{1},\dots,e_{n}$ for the
fibre and set $\<e_{i},e_{j}\>=a_{ij}$ where the matrix
$[a_{ij}]$ is positive definite; we should do this continuous in
every fiber. We get another fibration where the fiber is the set
of all positive-definite Hermitian matrices. We should take a
section of this fibration to get an inner product on each $F$,
to get a metric on the original fiber bundle. BUT the set of all
positive definite Hermitian matrices is contractible, by the
simple reason: given two inner products $\<-,-\>_{1}$ and
$\<-,-\>_{2}$ we can take their linear combination
$\alpha\<e,f\>_{1}+\beta\<e,f\>_{2}$ where $0\leq\alpha$,
$0\leq\beta$. This combination is another inner product. But
look, set $\beta=1-\alpha$ and we have a convex set which is
contractible. It is important that the inner product is positive,
otherwise we could end up with our linear combination resulting
in a degenerate inner product.
\end{ddanger}

\begin{lem}\index{Vector Bundle!Embedded into Trivial Bundle}\index{Trivial Bundle}
Every vector bundle may be embedded into a trivial bundle.
\end{lem}
The map of $E_1\to E_2$ is linear on the fibres, and an embedding
means that the map is injective on the fibres
\begin{equation}
\begin{array}{ccc}
E_{1} & \into & E_{2} \\
%\dTo  &       & \dTo \\
\downarrow & & \downarrow\\
B     & \xrightarrow{\id{}} & B
\end{array}
\end{equation}
How to construct such a map which is linear on the fibres?
Suppose that both are direct products
\begin{equation}
E_{1}=B\times F_1\quad\mbox{and}\quad E_2=B\times F_2
\end{equation}
then our diagram becomes
\begin{equation}
\begin{diagram}[small]
B\times F_{1} & & B\times F_{2}\\
\dTo>{p_{1}} & & \dTo>{p_{2}} \\
B & \rTo^{\id{}} & B.
\end{diagram}
\end{equation}
How do we construct a linear map on the fibres? Let $b\in B$ and
$\varphi_b\colon F_1\to F_2$ be the linear map we want. So
$\varphi_b\in\mathrm{Emb}(F_1,F_2)$. If $\dim(F_{1})=k$ and
$\dim(F_{2})=n$, then really $\mathrm{Emb}(F_1,F_2)=V_{n,k}$. So
if we had direct products, we'd need a continuous $B\to V_{n,k}$
to get an embedding of bundles.

Assume now that $B$ is a cell complex, the vector bundle is
trivial on each cell. Then we may apply this theory of
obstructions and we construct this map on some skeleton of
dimension $l$. Construct a map to a trivial bundle
\begin{equation}
B\times\CC^n\to B
\end{equation}
on $B^l$ (the $l$-dimensional skeleton), then extend it to
$B^{l+1}$. We may get an obstruction extending it to $B^{l+1}$
which would live in $H^{\bullet}\bigl(B,\pi_{l}(V_{n,k})\bigr)$.
If there is an obstruction, we go from 
\begin{equation}
n\to n+(\mbox{something})
\end{equation}
and when the (something) is large the obstruction disappears. We
can change $n$ since we introduced it in the skeleton approach
with the introduction of the trivial bundle $B\times\CC^{n}$
where $n$ was \emph{never} specified. So by taking
$n\mapsto n+(\mbox{something big})$, we remove the obstruction.

For every vector bundle $(E,B,F,p)$ we can find another vector
bundle $(E',B,F',p')$ that the direct sum of these two vector
bundles is a trivial vector bundle denoted by $\varepsilon^{N}$
(where $N$ is the dimension of the fiber). We may note that we
can talk about these guys as $\U{n}$ bundles, since we can always
introduce an inner product. Then we may always embed
$E\propersubset B\times\CC^{N}$ and everything we considered may
be equipped with an inner product. So we may consider
$E'=E^{\bot}$, and $\CC^{N}=F_{b}\oplus F_{b}^{\bot}$.

Now we can go to the definition of the $K$-group, which is now
very simple. Recall the notion of stable equivalence $E\stabequiv
E'$ if $E'\homotopic E+(\mbox{trivial bundle})$. We would now
like to say that bundles over $B$ with respect to stable
equivalence form a group $\widetilde{K}(B)$.

The operation is simply the direct sum, it is
well-defined. Inverses exist, we just proved this! The trivial
bundle is stable equivalent to 0, so the identity exists. It
forms a group.
\begin{ex}
If $B$ is a point, $\widetilde{K}(B)=0$ since everything is trivial.
\end{ex}
\begin{ex}
If $B=S^{n}$, then $\widetilde{K}(B)=\pi_{n-1}\bigl(\U{k}\bigr)$
if the fibres are of dimension $k$; we may consider any value of
$k$, so really we have $E\stabequiv E+\varepsilon^{N}$, so we may
consider a map
$\pi_{n-1}\bigl(\U{k}\bigr)\to\pi_{n-1}\bigl(\U{k+\mbox{something}}\bigr)$.

There is the stabilization of the group, we recall the notion of
$\U{\infty}$\index{$\U{\infty}$} such that 
\[
\pi_{n-1}\bigl(\U{\infty}\bigr)=\pi_{n-1}\bigl(\U{N}\bigr)
\]
where $N$ is ``sufficiently large.'' This is the answer
$\widetilde{K}(S^{n})=\pi_{n-1}\bigl(\U{\infty}\bigr)$. 
\end{ex}


